Anderson Video - Hydroelectric Dam

Professor Anderson
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>> Hello class, Professor Anderson here. Let's take a look at one of your homework problems. This is the problem of the hydroelectric dam. And so in this problem they say the following. You have water that is flowing over a dam and when it gets to the bottom, it hits a paddlewheel. Okay, and that paddlewheel rotates and creates electricity. So this is a hydroelectric dam, this is exactly how it works. And the idea is that we're going to take gravitational potential energy, we're going to convert that into electrical power, electrical energy. And we'll see how it works. So, a first question here is, let's say we have water falling of height h, what is the change in gravitational potential energy? So delta u is simply u final minus u initial. And since we're dealing with gravitational potential energy, this is equal to mg y final minus mg y initial. All right, mg factors out. We can write it as mg times the quantity y final minus y initial. And y final is at the bottom, it's 0, y initial is whatever they give us, and so it is minus mgh. Okay, that's the change in gravitational potential energy. Now all of that water falling on the paddlewheel makes it turn and you can make electricity that way. And so let's see if we can calculate how much water needs to fall to generate a certain amount of electricity. Okay, so to do that, we need to think about how much power we want to generate. Now a typical hydroelectric dam maybe generates 45 megawatts of power. Power is what? Power, remember, is work over time, okay. How much work did gravity do on the water falling? Well, there's a force, mg, it went at height, h, and so it's just the negative of the change in that potential energy. And so it is Mgh and it's falling in a time, t. Now I used a capital "M" here because 45 megawatts sounds like a lot of power and I suspect that 1 kilogram of water is not going to be able to do that, probably a lot of kilograms of water falling per second are required. And so that's why I used a capital "M" here. So what they are looking for is, what is the mass of water per second to generate that much power? And so if I look at this previous equation right here, M over t is just equal to p divided by gh. And now we have all those numbers, all right. So you can plug them in and try it, but there's a problem. The problem is this is assuming that our paddlewheel is 100% efficient. In other words, all the available energy in that falling water in fact gets turned into electricity, and we know that's not the case. No real system is 100% efficient. And so we have to include an efficiency factor. And so where do we include the efficiency factor? Well, if epsilon is our efficiency factor and let's say it's 80%, that is 0.8. And if I multiply this thing by epsilon, then I would get a smaller mass of water, but we know that can't be the case. We must have to use more water in order to get this amount of power. And so we have to, in fact, divide by epsilon. So what's the real answer here? It is p over epsilon gh. And now if you plug in those numbers, you'll get an answer that will make sense. It will be a big number and it will be kilograms of water per second, okay. Good luck with that one. If you have any questions, come see me in office hours. Cheers.
>> Hello class, Professor Anderson here. Let's take a look at one of your homework problems. This is the problem of the hydroelectric dam. And so in this problem they say the following. You have water that is flowing over a dam and when it gets to the bottom, it hits a paddlewheel. Okay, and that paddlewheel rotates and creates electricity. So this is a hydroelectric dam, this is exactly how it works. And the idea is that we're going to take gravitational potential energy, we're going to convert that into electrical power, electrical energy. And we'll see how it works. So, a first question here is, let's say we have water falling of height h, what is the change in gravitational potential energy? So delta u is simply u final minus u initial. And since we're dealing with gravitational potential energy, this is equal to mg y final minus mg y initial. All right, mg factors out. We can write it as mg times the quantity y final minus y initial. And y final is at the bottom, it's 0, y initial is whatever they give us, and so it is minus mgh. Okay, that's the change in gravitational potential energy. Now all of that water falling on the paddlewheel makes it turn and you can make electricity that way. And so let's see if we can calculate how much water needs to fall to generate a certain amount of electricity. Okay, so to do that, we need to think about how much power we want to generate. Now a typical hydroelectric dam maybe generates 45 megawatts of power. Power is what? Power, remember, is work over time, okay. How much work did gravity do on the water falling? Well, there's a force, mg, it went at height, h, and so it's just the negative of the change in that potential energy. And so it is Mgh and it's falling in a time, t. Now I used a capital "M" here because 45 megawatts sounds like a lot of power and I suspect that 1 kilogram of water is not going to be able to do that, probably a lot of kilograms of water falling per second are required. And so that's why I used a capital "M" here. So what they are looking for is, what is the mass of water per second to generate that much power? And so if I look at this previous equation right here, M over t is just equal to p divided by gh. And now we have all those numbers, all right. So you can plug them in and try it, but there's a problem. The problem is this is assuming that our paddlewheel is 100% efficient. In other words, all the available energy in that falling water in fact gets turned into electricity, and we know that's not the case. No real system is 100% efficient. And so we have to include an efficiency factor. And so where do we include the efficiency factor? Well, if epsilon is our efficiency factor and let's say it's 80%, that is 0.8. And if I multiply this thing by epsilon, then I would get a smaller mass of water, but we know that can't be the case. We must have to use more water in order to get this amount of power. And so we have to, in fact, divide by epsilon. So what's the real answer here? It is p over epsilon gh. And now if you plug in those numbers, you'll get an answer that will make sense. It will be a big number and it will be kilograms of water per second, okay. Good luck with that one. If you have any questions, come see me in office hours. Cheers.